## Algebra 2 (1st Edition)

$a_n= -3 \times (-5)^{n-1}$ $a_{15}=-18,310, 546, 875$ $S_{15}=-15,258, 789, 063$
The general formula for the nth term of a geometric series is given by $a_n= a_1r^{n-1}$ ...(1) The ratio of the successive terms is $r=-5$ . Equation (1) gives: $a_n= a_1 \times (-5)^{n-1}=-3 \times (-5)^{n-1}$ Plugging in $n =15$, we have $a_{15}=-3 \times (-5)^{14}=-18,310, 546, 875$ We know that $S_{n}=\dfrac{a_1(1-r^n)}{1-r}$ Now, $S_{15}=\dfrac{-3 \times (1-(-5)^{15})}{1-(-5)} =-15,258, 789, 063$